Reliability HotWire: eMagazine for the Reliability Professional
Reliability HotWire

Issue 11, January 2002

Reliability Basics

Nonparametric Analysis

In the last several issues of Reliability HotWire, we have looked at various ways of performing parameter estimations on life data sets. Nonparametric analysis allows the analyst to characterize life data without assuming an underlying distribution. This can be advantageous because it avoids some of the dangers inherent in an analysis where an inappropriate distribution is assumed. However, the analysis is limited to reliability estimates only for the failure times in the data set, thus making it impossible to make reliability predictions outside the range of data values. While this somewhat limits the usefulness of this method, nonparametric analysis can be used as a benchmark against which other parametric analyses can be compared for accuracy. ReliaSoft's Weibull++ software provides a nonparametric analysis utility.

There are several methods for conducting a nonparametric analysis, including the Kaplan-Meier, simple actuarial and standard actuarial methods. A method for attaching confidence bounds to the results of these nonparametric analysis techniques can also be developed. The basis of nonparametric life data analysis is the empirical cdf function, which is given by:

Empirical cdf function

Kaplan-Meier Estimator
The Kaplan-Meier estimator, also known as the product limit estimator, can be used to calculate values for nonparametric reliability for data sets with multiple failures and suspensions. The equation of the estimator is given by:

Equation of the Kaplan-Meier estimator

where m is the total number of data points and n is the total number of units under analysis. The variable ni is defined by:

Definition of the variable n sub i

where rj is the number of failures in the jth data group, and sj is the number of suspensions in the jth data group. Note that the reliability estimate is only calculated for times at which one or more failures occurred. For the sake of calculating the value of ni at time values that have failures and suspensions, it is assumed that the suspensions occur slightly after the failures, so that the suspended units are considered to be operating and included in the count of ni.

Kaplan-Meier Estimator Example
A group of 20 units are put on a life test with the following results.

Number in State State (F or S) State End Time
3 F 9
1 S 9
1 F 11
1 S 12
1 F 13
1 S 13
1 S 15
1 F 17
1 F 21
1 S 22
1 S 24
1 S 26
1 F 28
1 F 30
1 S 32
2 S 35
1 S 39
1 S 41

Using the data and the Kaplan-Meier estimator, the following table of reliability estimates can be constructed.

State End Time Number of Failures, ri Number of Suspensions, si Available Units, ni

9 3 1 20 0.850 0.850
11 1 0 16 0.938 0.797
12 0 1 15 1.000 0.797
13 1 1 14 0.929 0.740
15 0 1 12 1.000 0.740
17 1 0 11 0.909 0.673
21 1 0 10 0.900 0.605
22 0 1 9 1.000 0.605
24 0 1 8 1.000 0.605
26 0 1 7 1.000 0.605
28 1 0 6 0.833 0.505
30 1 0 5 0.800 0.404
32 0 1 4 1.000 0.404
35 0 1 3 1.000 0.404
39 0 1 2 1.000 0.404
41 0 1 1 1.000 0.404

 As can be determined from the preceding table, the reliability estimates for the failure times are:

Failure Time Reliability Estimate
9 85.0%
11 79.7%
13 74.0%
17 67.3%
21 60.5%
28 50.5%
30 40.4%

Simple Actuarial Method
The simple actuarial method is an easy-to-use form of nonparametric data analysis that can be used for multiply censored data that are arranged in intervals. This method is based on calculating the number of failures in a time interval, rj, versus the number of operating units in that time period, ni. The equation for the reliability estimator for the simple actuarial method is given by:

Equation for the simple actuarial estimator

where m is the total number of intervals and n is the total number of units under analysis. The variable ni is defined by:

Definition of the variable n sub i

where rj is the number of failures in the jth data group and sj is the number of suspensions in the jth data group.

Simple Actuarial Method Example
A group of 55 units are put on a life test during which the units are evaluated every 50 hours. The test had the following results:

Start Time End Time Number of Failures, ri Number of Suspensions, si
0 50 2 4
50 100 0 5
100 150 2 2
150 200 3 5
200 250 2 1
250 300 1 2
300 350 2 1
350 400 3 3
400 450 3 4
450 500 1 2
500 550 2 1
550 600 1 0
600 650 2 1

The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in the calculation of the reliability estimates for the simple actuarial equation.

Start Time End Time Number of Failures, ri Number of Suspensions, si Available Units, ni

0 50 2 4 55 0.964 0.964
50 100 0 5 49 1.000 0.964
100 150 2 2 44 0.955 0.920
150 200 3 5 40 0.925 0.851
200 250 2 1 32 0.938 0.798
250 300 1 2 29 0.966 0.770
300 350 2 1 26 0.923 0.711
350 400 3 3 23 0.870 0.618
400 450 3 4 17 0.824 0.509
450 500 1 2 10 0.900 0.458
500 550 2 1 7 0.714 0.327
550 600 1 0 4 0.750 0.245
600 650 2 1 3 0.333 0.082

As can be determined from the preceding table, the reliability estimates for the failure times are:

Failure Period
End Time
Reliability
Estimate
50 96.4%
150 92.0%
200 85.1%
250 79.8%
300 77.0%
350 71.1%
400 61.8%
450 50.9%
500 45.8%
550 32.7%
600 24.5%
650 8.2%

Standard Actuarial Method
The standard actuarial model is a variation of the simple actuarial model that involves adjusting the value for the number of operating units in an interval. The Kaplan-Meier and simple actuarial methods assume that the suspensions in a time period or interval occur at the end of that interval, after the failures have occurred. The standard actuarial model assumes that the suspensions occur in the middle of the interval, which has the effect of reducing the number of available units in the interval by half of the suspensions in that interval or:

With this adjustment, the calculations are carried out just as they were for the simple actuarial model, or:

Standard Actuarial Method Example
In this example, we will apply the standard actuarial method to the data set used in the simple actuarial method example. The solution to this example is similar to that of the previous example, with the exception of the inclusion of the ni' term, which is used in the preceding equation. Applying this equation to the data, we can generate the following table:

Start Time End Time Number of Failures, ri Number of Suspensions, si Adjusted Units, ni'

0 50 2 4 53 0.962 0.962
50 100 0 5 46.5 1.000 0.962
100 150 2 2 43 0.953 0.918
150 200 3 5 37.5 0.920 0.844
200 250 2 1 31.5 0.937 0.791
250 300 1 2 28 0.964 0.762
300 350 2 1 25.5 0.922 0.702
350 400 3 3 21.5 0.860 0.604
400 450 3 4 15 0.800 0.484
450 500 1 2 9 0.889 0.430
500 550 2 1 6.5 0.692 0.298
550 600 1 0 4 0.750 0.223
600 650 2 1 2.5 0.200 0.045

As can be determined from the preceding table, the reliability estimates for the failure times are:

Failure Period
End Time
Reliability
Estimate
50 96.2%
150 91.8%
200 84.4%
250 79.1%
300 76.2%
350 70.2%
400 60.4%
450 48.4%
500 43.0%
550 29.8%
600 22.3%
650 4.5%

Nonparametric Confidence Bounds
Confidence bounds for nonparametric reliability can be calculated in a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the for nonparametric data, the Weibull++ software uses Greenwood's formula:

Greenwood's formula

where m is the total number of intervals and n is the total number of units under analysis. The variable ni is defined by:

Definition of the variable n sub i

where rj is the number of failures in the jth data group and sj is the number of suspensions in the jth data group. Once the variance has been calculated, the standard error can be determined by taking the square root of the variance,

Equation for the square root of the variance

This information can then be applied to determine the confidence bounds,

Equation for confidence bounds

where:

and α is the desired confidence level for the 1-sided confidence bounds.

Nonparametric Confidence Bounds Example
In this example, we will determine the 1-sided confidence bounds for the reliability estimates in the standard actuarial method example above, with a 97.5% confidence level. Once again, this type of problem is most readily solved by constructing a table similar to the following:

Failure Time Reliability Estimate Number of Failures Adjusted Units ri/ni' Variance Error w Lower CL Upper CL
50 0.962 2 53 0.0377 0.0007 0.0262 4.108 0.861 0.991
150 0.918 2 43 0.0465 0.0016 0.0397 2.797 0.799 0.969
200 0.844 3 37.5 0.0800 0.0030 0.0547 2.257 0.706 0.924
250 0.791 2 31.5 0.0635 0.0040 0.0630 2.107 0.642 0.888
300 0.762 1 28 0.0357 0.0045 0.0668 2.059 0.609 0.868
350 0.702 2 25.5 0.0784 0.0054 0.0737 1.996 0.542 0.825
400 0.604 3 21.5 0.1395 0.0068 0.0823 1.964 0.438 0.750
450 0.484 3 15 0.2000 0.0082 0.0907 2.038 0.315 0.656
500 0.430 1 9 0.1111 0.0091 0.0953 2.142 0.260 0.618
550 0.298 2 6.5 0.3077 0.0104 0.1020 2.602 0.140 0.524
600 0.223 1 4 0.2500 0.0100 0.1000 3.098 0.085 0.471
650 0.045 2 2.5 0.8000 0.0036 0.0599 15.69 0.003 0.423

The following plot illustrates these results graphically:

Plot for the nonparametric analysis example with confidence bounds

 

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